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Math Help - Dice Probability

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    Dice Probability

    A fair n-sided die is rolled n times. Assuming the rolls are independent, calculate the probability of getting a match on roll i, i.e. on roll i the die shows i.
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    Quote Originally Posted by gtaplayr View Post
    A fair n-sided die is rolled n times. Assuming the rolls are independent, calculate the probability of getting a match on roll i, i.e. on roll i the die shows i.
    I dont think you have posted what you are really asking.
    Suppose we roll an ordinary die six times.
    The probability of getting a 1 on the first roll is \frac{1}{6}.
    The probability of getting a 2 on the second roll is \frac{1}{6}.

    The probability of getting a 6 on the sixth roll is \frac{1}{6}.

    Now is that truly what you asked?
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    Yes i understand that the likelihood of a matct is 1/n on any particular trial.
    But i dont understand how to calulate the probability of getting a match on roll i.
    Where does n start and where does it end?
    If its infinite then isnt the probabilty of getting a match 1?
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    Quote Originally Posted by gtaplayr View Post
    Yes i understand that the likelihood of a matct is 1/n on any particular trial.
    But i dont understand how to calulate the probability of getting a match on roll i.
    Where does n start and where does it end?
    If its infinite then isnt the probabilty of getting a match 1?
    Suppose we roll an ordinary die six times.
    If you want the probability if getting the string 123456 then that is \left(\frac{1}{6}\right)^6.

    For an n-sided die rolled n times, the probability if getting the string 123n is \left(\frac{1}{n}\right)^n.
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    Quote Originally Posted by gtaplayr View Post
    A fair n-sided die is rolled n times. Assuming the rolls are independent, calculate the probability of getting a match on roll i, i.e. on roll i the die shows i.
    Perhaps you mean what is the probability of getting at least one match?

    If so, this is 1 - P(no match). The probability of not getting a match on the nth roll is 1-1/n, so assuming the rolls are independent, the probability of no macth in n rolls is (1-1/n)^n, and

    P(\text{at least one match}) = 1 - (1 - 1/n)^n.

    If you are interested in large n,

    \lim_{n \to \infty} 1 - (1 - 1/n)^n = 1 - 1/e.
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